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A worker carries two bags. Each of the bags initially contains N nails. Whenever the worker needs a nail, he takes it from a bag picked at random. At some point the worker picks an empty bag. Find the probability that the other bag contains exactly m nails.

My reasoning:

The desired probability: $C_2^1 \cdot P(|A| = 0\ and\ |B| = m)$, where A, B - sets of nails in each bag. Then we can "divide" this probability (I suppose that I deal with independent events):

$P(|A| = 0\ and\ |B| = m) = P(|A| = 0) \cdot P(|B| = m);$

$P(|A| = 0) = (\frac{1}{2})^N$, where $|A| = |B| = N$ (initial condition).

$P(|B| = m) = (\frac{1}{2})^{N - m}$

Result: $C_2^1 \cdot P(|A| = 0\ and\ |B| = m)$ = $C_2^1 \cdot (\frac{1}{2})^{2N - m}$

Am I wrong or it is correct?

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An easier way to think about it is like this: consider picking bag $1$ a success and bag $2$ a failure. The probability of bag $1$ being empty and bag $2$ having $m$ nails is the probability of $N$ successes and $N-m$ failures. This is an easy binomial calculation: ${2N-m\choose N}\frac{1}{2^{2N-m}}$. For the next pick, we want the worker to choose the empty bag, so we multiply that by $\frac{1}{2}$ to get ${2N-m\choose N}\frac{1}{2^{2N-m+1}}$. However the reverse scenario also counts (bag $1$ having $m$ and bag $2$ being empty). This amounts to $N-m$ successes and $N$ failures. By the same logic, we get ${2N-m\choose N-m}\frac{1}{2^{2N-m+1}}$ Add these two possibilities together and we get ${2N-m\choose N}\frac{1}{2^{2N-m+1}}+{2N-m\choose N-m}\frac{1}{2^{2N-m+1}}$

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  • $\begingroup$ Ah, I'm an idiot, thanks. Hopefully fixed it. $\endgroup$
    – norvia
    May 16 '20 at 23:02
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    $\begingroup$ Just one more thing. The given sum simplifies to $$\binom{2n-m}{n} 2^{m-2n}.$$ $\endgroup$
    – heropup
    May 16 '20 at 23:09
  • $\begingroup$ The first and the second situations are identical, so we can just multiply by 2 $\endgroup$
    – tikhpavel
    Sep 28 at 15:56

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